A vertical tower stands on a horizontal ground. Two points P and Q are located on the ground due East and due North of the tower, respectively. The angles of elevation of the top of the tower from P and Q are 45^ and 30^ , respectively. If the distance between P and Q is 40 meters, then the height of the tower (in meters) is

Step-by-step solution

Let the height of the tower be h and its base be O . The distance of point P from the base of the tower is OP = h 45^ = h . The distance of point Q from the base of the tower is OQ = h 30^ = h 3 . Since P is due East and Q is due North of the tower, the angle POQ is 90^ . In the right-angled triangle POQ , applying the Pythagoras theorem: PQ^2 = OP^2 + OQ^2 Substitute the given distance PQ = 40 m: 40^2 = h^2 + (h 3 )^2 1600 = h^2 + 3h^2 1600 = 4h^2 h^2 = 400 h = 20 Thus, the height of the tower is 20 meters. Answer